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Context: Computational Chemistry

So, I am looking at the molecular orbital approximation, which allows the assumption that the overall wavefunction can be written as individual molecular orbitals

Ψ(r1,r2..,rn) = ψ1(r1) x ψ2(r2) x .... ψn(rn) (the numbers and n are subscripted)

However, the above equation is incorrect because it does not take into account spin. Therefore, when spin is taken into account, the molecular orbital ψ(r) is replaced by molecular spin orbitals χ(r,ω). The r and ω are often combined and denoted as xi (the i is subscripted). Does this mean that χ = ψ? If that is true, I don't understand why the symbol was changed from ψ to χ? If someone could explain. Thank you.

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    $\begingroup$ "the numbers and n are subscripted" — please take a look at FAQ: How can I format math/chemistry expressions on Chemistry Stack Exchange?. That said, it's quite hard, perhaps even impossible, to answer your question. Where did you get this equation from? What is the context within which it was written? It is very possible that the author is just making a simplification (albeit an incorrect one) where spin is ignored. But we can't really read their mind; and certainly not when we only have one equation of their writing to work with. $\endgroup$ Commented Mar 17, 2023 at 1:23
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    $\begingroup$ (Like, sure, you said it's about comp chem and the MO approximation; but that aspect of it is already quite clear from the equation you quoted, at least to anybody who's studied it before. The real question is: what did the author write before and after the equation? In their article or book, have they consistently ignored the spin degrees of freedom, using only $\psi$'s (spatial orbitals) instead of $\chi$'s (spin orbitals)? Or did they actually cover spin, and are just using a different notation from what you're expecting? These are all possible answers; but it's impossible to say which.) $\endgroup$ Commented Mar 17, 2023 at 1:28
  • $\begingroup$ Hello, thank you for linking the FAQ. I'll use it in the not too distant future. $\endgroup$ Commented Mar 18, 2023 at 1:36
  • $\begingroup$ Hello, thank you for linking the FAQ. I'll use it in the not too distant future. The book I am reading, Computational Chemistry by Jeremy Harvey (Oxford Primer), is where I found this equation. The section of the book this equation is from is introducing the Hartree-Fock theory. In the build up to the equation, spin is ignored. Shortly after the introduction of the equation, the author states the equation is not quite right as it ignores spin. Then the spin coordinate, ω, and spin part is introduced, with spin up and down. Then ($r_i$, $ω_i$) = $x_i$ $\endgroup$ Commented Mar 18, 2023 at 1:58
  • $\begingroup$ ($r_i$, $ω_i$) = $x_i$ Is the combination of the spin and spatial variable for a given electron. A few statements are then made. The first is about how electrons are need to be treated as identical particles. The second is about swapping electrons being antisymmetric. As a result of these statements and the spin terminology introduced, the equation in the question becomes ψ($x_1$,$x_2$,$x_3$,$x_4$)=-ψ($x_1$,$x_3$,$x_2$,$x_4$) for a 4 electron system. $\endgroup$ Commented Mar 18, 2023 at 2:07

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I think maybe there is some confusion in terminology. Maybe these notes will help you:

  1. When we take spin into account, we combine the spatial orbital of the ith electron $ \psi_i(\mathbf{r}) $ that depends on the position with the spin function (of course we have two of them), $ \alpha(\omega) $ or $ \beta(\omega)$ that depend on the spin coordinate. Formally we don't combine $ \mathbf{r} $ and $ \omega $, we take the product of $ \psi_i(\mathbf{r}) $ and $ \alpha(\omega) $ or $ \beta(\omega) $. However, we combine these two coordinates in $ \mathbf{x} = (\mathbf{r},\omega) $.
  2. The product of this new function is the spin orbital $ \chi(\mathbf{x}) = \psi(\mathbf{r})\alpha(\omega) $ or $ \chi(\mathbf{x}) = \psi(\mathbf{r})\beta(\omega) $ .
  3. What we want to calculate is the N-electron wave function, this is the guy you wrote there in the left-hand side $ \Phi(\mathbf{x}_1, ...,\mathbf{x}_N) $. However, when I read the title of the question you talk about the Slater determinant, maybe you wanted to write the following \begin{equation} \Phi(\mathbf{x}_1, ...,\mathbf{x}_N) = \frac{1}{\sqrt{N!}} \det \begin{bmatrix}{} \chi_i(\mathbf{x}_1) & \chi_j(\mathbf{x}_1) & \cdots & \chi_k(\mathbf{x}_1) \\ \chi_i(\mathbf{x}_2) & \chi_j(\mathbf{x}_2) & \cdots & \chi_k(\mathbf{x}_2) \\ \vdots & \vdots & \vdots & \vdots \\ \chi_i(\mathbf{x}_N) & \chi_j(\mathbf{x}_N) & \cdots & \chi_k(\mathbf{x}_N) \\ \end{bmatrix} \end{equation}

So, in summary:

  1. When we take spin into account, we change $ \psi $ for $ \chi $.
  2. What goes in the Slater determinant is always $ \chi $.
  3. The N-electron wave function is the capital $ \Phi $.
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  • $\begingroup$ Hello, thank you very much. The second point answered my question. Also, thanks for the additional information on the Slater determinant. $\endgroup$ Commented Mar 18, 2023 at 1:32
  • $\begingroup$ My pleasure! I assume you are studying the Hartree-Fock method, it is tough, somewhat heavy math, but at the end is very, very rewarding... $\endgroup$ Commented Mar 18, 2023 at 20:42

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